2.1 Twisted Alexander polynomial

Let $K \subset S^3$ be a knot and $E_K= S^3 \setminus K$ the knot exterior. Let $G(K)=\pi _1(E_K)$ be the knot group, which is the fundamental group of $E_K$ . We choose a Wirtinger presentation:

$$ \begin{align*}G(K) = \langle x_1,\ldots,x_l\,|\,r_1,\ldots,r_{l-1}\rangle.\end{align*} $$

The abelianization homomorphism

$$ \begin{align*}\mathfrak{a}:G(K) \to H_1(E_K;\mathbb Z)\cong\mathbb Z = \langle t \rangle \end{align*} $$

is given by assigning to each generator $x_i$ the meridian element $t \in H_1(E_K;\mathbb Z)$ . Here, we denote the sum in $\mathbb Z$ multiplicatively.

Let $\rho :G(K)\to \mathrm {SL}_d(\mathbb C)$ be a $d$ -dimensional linear representation. The maps $\rho $ and $\mathfrak {a}$ naturally induce two ring homomorphisms $\tilde {\rho }: {\mathbb Z}[G(K)] \rightarrow M_d(\mathbb C)$ and $\tilde {\mathfrak {a}}:{\mathbb Z}[G(K)]\rightarrow {\mathbb Z}[t^{\pm 1}]$ , where ${\mathbb Z}[G(K)]$ is the group ring of $G(K)$ and $M_d(\mathbb C)$ is the matrix algebra of degree $d$ over ${\mathbb C}$ . The tensor product $\tilde {\rho }\otimes \tilde {\mathfrak {a}}$ defines a ring homomorphism ${\mathbb Z}[G(K)] \to M_d \left ({\mathbb C}[t^{\pm 1}]\right )$ . Let $F_l$ denote the free group on the generators $x_1,\ldots ,x_l$ and $\Phi :{\mathbb Z}[F_l]\to M_d\left ({\mathbb C}[t^{\pm 1}]\right )$ the composition of the surjection $\tilde {\phi }:{\mathbb Z}[F_l]\to {\mathbb Z}[G(K)]$ induced by the presentation of $G(K)$ and the map $\tilde {\rho }\otimes \tilde {\mathfrak {a}}:{\mathbb Z}[G(K)]\to M_d\left ({\mathbb C}[t^{\pm 1}]\right )$ .

Let $Q$ denote the $(l-1)\times l$ matrix whose $(i,j)$ -entry is the $d\times d$ matrix $\Phi \left (\frac {\partial r_i}{\partial x_j}\right ) \in M_d\left ({\mathbb C}[t^{\pm 1}]\right )$ , where $\frac {\partial }{\partial x}$ denotes the Fox derivative. For $1\leq j\leq l$ , we denote by $Q_j$ the $(l-1)\times (l-1)$ matrix obtained from $Q$ by removing the $j$ -th column. We regard $Q_j$ as a $d(l-1)\times d(l-1)$ matrix with coefficients in ${\mathbb C}[t^{\pm 1}]$ . Then Wada’s twisted Alexander polynomial [Reference Wada22] of $K$ associated with a representation $\rho : G(K) \to \mathrm {SL}_d(\mathbb C)$ is defined to be the rational function

$$ \begin{align*}\Delta_{K}^{\rho}(t) =\frac{\det Q_j}{\det\Phi(1-x_j)}. \end{align*} $$

It is well-defined up to multiplication by $\pm t^{i}~(i\in {\mathbb Z})$ .

Now, we focus on two-dimensional linear representations of knot groups. It is known that if $\rho :G(K)\to \mathrm {SL}_2(\mathbb C)$ is a non-abelian representation, the rational function $\Delta _{K}^{\rho }(t)$ becomes a Laurent polynomial in $t$ [Reference Kitano and Morifuji7, Theorem 3.1].

Let $\mathrm {sl}_2(\mathbb C)$ be the Lie algebra of $2 \times 2$ complex matrices with trace zero. The adjoint action, denoted by $\mathrm {Ad}$ , is the conjugation on $\mathrm {sl}_2(\mathbb C)$ by $\mathrm {SL}_2(\mathbb C)$ , that is, $\mathrm {Ad}(V)(g)=VgV^{-1} \in \mathrm {sl}_2(\mathbb C)$ for all $V \in \mathrm {SL}_2(\mathbb C)$ and $g \in \mathrm {sl}_2(\mathbb C)$ . For each representation $\rho : G(K) \to \mathrm {SL}_2(\mathbb C)$ , the composition $\mathrm {Ad} \circ \rho $ is a representation of $G(K)$ into $\mathrm {SL}_3(\mathbb C)$ and thus we can define a rational function $\Delta _K^{\mathrm {Ad} \circ \rho } (t)$ called the adjoint twisted Alexander polynomial of $K$ .

2.2 Hyperbolic torsion conjecture

For a hyperbolic knot $K$ in $S^3$ , recall that $\rho _{0}: G(K) \to \mathrm {SL}_2(\mathbb C)$ is a lift of the holonomy representation $\bar {\rho }_{0}: G(K) \to \mathrm {PSL}_2(\mathbb C)$ corresponding to the hyperbolic structure of the knot exterior. The representation $\rho _{0}$ is known to be discrete and faithful [Reference Thurston19]. In [Reference Dunfield, Friedl and Jackson3], Dunfield et al. studied the hyperbolic torsion polynomial $\mathcal T_K(t)$ which is a normalization of the twisted Alexander polynomial $\Delta _{K}^{\rho _{0}}(t) \in \mathbb C[t^{\pm 1}]$ such that $\mathcal T_K(t) = \mathcal T_K(t^{-1})$ . Based on many numerical computations, they conjectured that $\mathcal T_K(t) \in \mathbb C[t^{\pm 1}]$ determines the knot genus $g(K)$ , namely, $\deg \mathcal T_K(t)= 2(2g(K)-1)$ . Moreover, $K$ is fibered if and only if $\mathcal T_K(t)$ is monic. This conjecture has been verified for some families of hyperbolic knots [Reference Agol and Dunfield1, Reference Morifuji10, Reference Morifuji11, Reference Morifuji and Tran13, Reference Morifuji and Tran14, Reference Porti16].

On the other hand, we can also study the adjoint twisted Alexander polynomial associated with the holonomy representation. In [Reference Dubois and Yamaguchi2], Dubois and Yamaguchi proved that the twisted Alexander polynomial $\Delta _{K}^{\mathrm {Ad} \circ \rho _{0}}(t)$ is actually a Laurent polynomial in $t$ . The adjoint hyperbolic torsion polynomial $\mathcal T^{\mathrm {Ad}}_K(t) \in \mathbb C[t^{\pm 1}]$ is a normalization of $\Delta _{K}^{\mathrm {Ad} \circ \rho _{0}}(t)$ such that $\mathcal T^{\mathrm {Ad}}_K(t^{-1}) = - \mathcal T^{\mathrm {Ad}}_K(t)$ up to multiplication by $t^i \, (i \in \mathbb Z)$ . In [Reference Dunfield, Friedl and Jackson3], Dunfield et al. remarked that $\mathcal T^{\mathrm {Ad}}_K(t)$ does not always determine the knot genus. However, for all hyperbolic knots having at most $15$ crossings, they numerically checked that $K$ is fibered if and only if $\mathcal T^{\mathrm {Ad}}_K(t)$ is monic. In [Reference Morifuji12, Reference Tran21], by using an explicit formula for the adjoint twisted Alexander polynomial $\Delta _{K}^{\mathrm {Ad} \circ \rho }(t)$ from [Reference Tran20], Morifuji and the author proved that the adjoint hyperbolic torsion polynomial determines the genus and fibering of genus one two-bridge knots.

2.3 Algebraic integers

An algebraic integer is a complex number which is a root of some monic polynomial whose coefficients are integers. It is easily checked that the set of all algebraic integers, denoted by $\mathbb A$ , is closed under addition and multiplication. Hence, $\mathbb A$ is a commutative subring of $\mathbb C$ . We will show the following.

Proof The proof is similar to that of Eisenstein’s irreducibility criterion.

Suppose $t_0$ is a root of $P(t)$ . Let $R(t)$ be the minimal polynomial of $t_0$ over $\mathbb Z$ . Write $R(t) = a_r t^r + \cdots + a_1 t + a_0 \in \mathbb Z[t]$ . We claim that $p \mid a_r$ .

Note that $R \mid P$ . If $\deg R = \deg P$ , then $P= d R$ for some $d \in \mathbb Z$ . Since $p \nmid c_0= da_0$ , we have $p \nmid d$ . Then $p \mid c_q = da_r$ implies that $p \mid a_r$ .

Suppose $\deg R < \deg P$ then $P(t) = R(t) S(t)$ for some non-constant polynomial $S(t) \in \mathbb Z[t]$ . Write $S(t) = b_s t^s + \cdots + b_1 t + b_0$ . Note that $p \nmid c_0=a_0 b_0$ . Let $i \le r$ be the largest integer such that $p \nmid a_i$ , and $j \le s$ the largest integer such that $p \nmid b_j$ . Consider

$$ \begin{align*}c_{i+j} = \sum_{u+v=i+j} a_u b_{v} = a_i b_j + \sum_{\substack{u+v=i+j, \\ (u,v) \not= (i,j)}} a_u b_{v}.\end{align*} $$

If $u>i$ then $p \mid a_u$ . If $u<i$ then $v=i+j-u>j$ and so $p \mid b_v$ . This implies that $p \mid a_u b_v$ if $u \not = i$ . Hence,

$$ \begin{align*}p \mid \sum_{\substack{u+v=i+j, \\ (u,v) \not= (i,j)}} a_u b_{v}. \end{align*} $$

Since $p \nmid a_ib_j$ , we obtain $p \nmid c_{i+j}$ . This only occurs when $i+j=0$ , namely, ${i=j=0}$ . Hence, $p \mid a_u$ for all $1 \le u \le r$ . In particular, $p \mid a_r$ . This implies that $a_r \not = \pm 1$ and therefore $t_0$ cannot be an algebraic integer.

3.1 The Riley polynomial

A representation $\rho : G({\mathcal {K}}) \to \mathrm {SL}_2(\mathbb C)$ is called non-abelian if its image $\rho (G({\mathcal {K}}))$ is a non-abelian subgroup of $\mathrm {SL}_2(\mathbb C)$ . Taking conjugation if necessary, we can assume that $\rho $ has the form

(3.1) $$ \begin{align} \rho(a) = \left[ \begin{array}{cc} M & 1 \\ 0 & M^{-1} \end{array} \right] \quad \text{and} \quad \rho(b) = \left[ \begin{array}{cc} M & 0 \\ 2-y & M^{-1} \end{array} \right], \end{align} $$

where $(M, y) \in \mathbb C^2$ satisfies $\rho (w^n a) = \rho (bw^n)$ . Note that $y = {\mathrm {tr}} \rho (ab^{-1})$ . By [Reference Riley18], the matrix equation $\rho (w^n a) = \rho (bw^n)$ is equivalent to a single polynomial equation $R_{\mathcal {K}}(x,y)=0$ , where $x := {\mathrm {tr}} \rho (a) = {\mathrm {tr}} \rho (b) = M+M^{-1}$ and $R_{\mathcal {K}}(x, y):=\rho (w^n)_{11} + (M^{-1} - M) \rho (w^n)_{12} \in \mathbb {C}[x, y]$ is the Riley polynomial of the two-bridge knot ${\mathcal {K}}$ . Here, $\rho (w^n)_{ij}$ is the $ij$ -th entry of the matrix $\rho (w^n)$ .

To determine the Riley polynomial of ${\mathcal {K}} = J(k,2n)$ , we first compute $\rho (w)$ via Chebyshev polynomials. Let $S_l(v), \, l \in \mathbb Z,$ be the Chebyshev polynomials of the second kind defined by $S_0(v)=1$ , $S_1(v)=v$ , and $S_{l}(v) = v S_{l-1}(v) - S_{l-2}(v)$ for all integers $l$ .

The following properties of $S_l(v)$ ’s are well-known (see, e.g., [Reference Morifuji and Tran14]).

We will also use the following result whose proof is based on the Cayley–Hamilton theorem $V^2 = ({\mathrm {tr}} V) V - ( \det V)I$ for $2 \times 2$ matrices $V \in M_2(\mathbb C)$ , where $I$ denotes the identity $2 \times 2$ matrix (see, e.g., [Reference Morifuji and Tran13]).

Lemma 3.2 Let $V \in \mathrm {SL}_2(\mathbb C)$ and $v={\mathrm {tr}} V$ . Then

$$ \begin{align*}V^l = S_l(v) I - S_{l-1}(v) V^{-1} = S_{l-1}(v) V - S_{l-2}(v) I \end{align*} $$

for any integer $l$ .

Then the matrix $\rho (w)$ can be computed explicitly as follows.

Proposition 3.3 $(1)$ If $k=2m$ , then

$$ \begin{align*} \rho(w) =& I - S_m(y) S_{m-1}(y) \left[ \begin{array}{cc} y-2 & M - M^{-1} \\ (M-M^{-1})(2-y) & 2-y \end{array} \right]\\ & + \, S^2_{m-1}(y) \left[ \begin{array}{cc} (y-2)(y-M^2) & - (M^{-1} y - M - M^{-1}) \\ (y-2)(M^{-1}y - M - M^{-1}) & M^{-2}(2-y) \end{array} \right]. \end{align*} $$

$(2)$ If $k=2m+1$ , then

$$ \begin{align*} \rho(w) &= I - S_m(y) S_{m-1}(y) \left[ \begin{array}{cc} 2M^2-y & M + M^{-1} \\ (M+M^{-1}) (2-y)& 2M^{-2}-y \end{array} \right] \\ & + \, S^2_m(y) \left[ \begin{array}{cc} M^2-1 & M \\ M (2-y) & -y + 1 + M^{-2} \end{array} \right] + S^2_{m-1}(y) \left[ \begin{array}{cc} -y+1 + M^2 & M^{-1} \\ M^{-1}(2-y) & M^{-2}-1 \end{array} \right]. \end{align*} $$

Proof $(1)$ If $k=2m$ , then $w = (ba^{-1})^m(b^{-1}a)^m$ . Since ${\mathrm {tr}} \rho (ba^{-1}) = {\mathrm {tr}} \rho (b^{-1}a) = y$ , by Lemma 3.2, we have

$$ \begin{align*} \rho(w) &= \left( S_m(y) I - S_{m-1}(y) \rho(ab^{-1}) \right) \left( S_m(y) I - S_{m-1}(y) \rho(a^{-1}b) \right) \\ &= S^2_m(y) I - S_m(y) S_{m-1}(y) \left( \rho(ab^{-1}) + \rho(a^{-1}b) \right)+ S^2_{m-1}(y) \rho(ab^{-1}a^{-1}b). \end{align*} $$

By Lemma 3.1, we have $S^2_m(y) - y S_m(y) S_{m-1}(y) + S^2_{m-1}(y) = 1$ . Hence,

$$ \begin{align*}\rho(w) = I - S_m(y) S_{m-1}(y) \left( \rho(ab^{-1}) + \rho(a^{-1}b) - y I\right)+ S^2_{m-1}(y) \left( \rho(ab^{-1}a^{-1}b) - I \right). \end{align*} $$

$(2)$ If $k=2m+1$ , then $w = (ba^{-1})^mba(b^{-1}a)^m$ . Hence,

$$ \begin{align*} \rho(w) &= \left( S_m(y) I - S_{m-1}(y) \rho(ab^{-1}) \right) \rho(ba) \left( S_m(y) I - S_{m-1}(y) \rho(a^{-1}b) \right) \\ &= S^2_m(y) \rho(ba) - S_m(y) S_{m-1}(y) \left( \rho(a^2) + \rho(b^2) \right) + S^2_{m-1}(y) \rho(ab) \\ &= I - S_m(y) S_{m-1}(y) \left( \rho(a^2) + \rho(b^2) - y I\right) \\ & \quad + S^2_m(y) \left( \rho(ba) - I \right) + S^2_{m-1}(y) \left( \rho(ab) - I \right). \end{align*} $$

The lemma follows by using the matrix forms of $\rho (a)$ and $\rho (b)$ in (3.1).

Let $z: = {\mathrm {tr}} \rho (w) = \rho (w)_{11} + \rho (w)_{22}$ . By taking the trace of $\rho (w)$ in Proposition 3.3 and noting that $M^2+M^{-2} = x^2-2$ , we obtain

(3.2) $$ \begin{align} z = \begin{cases} 2 + (y-2)(y+2-x^2)S^2_{m-1}(y) & \text{if~} k=2m,\\ 2 - (y+2-x^2) \left( S_m(y) - S_{m-1}(y) \right)^2 & \text{if~} k=2m+1. \end{cases} \end{align} $$

We can now compute the Riley polynomial $R_{\mathcal {K}}(x,y) = \rho (w^n)_{11} + (M^{-1} - M)\rho (w^n)_{12}$ . By Lemma 3.2, we have $\rho (w^n) = S_{n-1}(z) \rho (w) - S_{n-2}(z) I$ . Hence, we can write $R_{\mathcal {K}} = t S_{n-1}(z) - S_{n-2}(z)$ , where $t = \rho (w)_{11} + (M^{-1} - M) \rho (w)_{12}$ .

If $k=2m$ , then by Proposition 3.3, we have

$$ \begin{align*} t &= 1 - S_m(y) S_{m-1}(y) \left( y-2 - (M-M^{-1})^2 \right) \\&\quad + \, S^2_{m-1}(y) \left( (y-2)(y-M^2) + (M-M^{-1}) (M^{-1} y - M - M^{-1}) \right) \\& = 1 - S_m(y) S_{m-1}(y) (y- M^2-M^{-2}) + S^2_{m-1}(y) (y-1)(y- M^2-M^{-2}). \end{align*} $$

If $k=2m+1$ , then it follows from Proposition 3.3 that

$$ \begin{align*} t &= 1 - S_m(y) S_{m-1}(y) \left( (2M^2-y )- (M - M^{-1}) (M + M^{-1}) \right) \\& \quad + S^2_{m-1}(y) \left( (-y+1 + M^2) - (M - M^{-1}) M^{-1} \right) \\&= 1 - S_m(y) S_{m-1}(y) (M^2+M^{-2} -y) + S^2_{m-1}(y) (M^2+M^{-2} -y). \end{align*} $$

Hence, $R_{\mathcal {K}}(x,y) = t S_{n-1}(z) - S_{n-2}(z)$ , where

(3.3) $$ \begin{align} t = \begin{cases} 1-(y+2-x^2)S_{m-1}(y)(S_m(y) - (y-1)S_{m-1}(y)) & \text{if~} k=2m,\\ 1+(y+2-x^2)S_{m-1}(y)(S_m(y) - S_{m-1}(y)) & \text{if~} k=2m+1. \end{cases} \end{align} $$

See [Reference Macasieb, Petersen and van Luijk9, Reference Morifuji and Tran13] for other ways to obtain (3.2) and (3.3).

Proof $(1)$ Suppose $k=2m$ . If $S_{m-1}(y)=0$ , then by (3.2) and (3.3), we have $z=2$ and $t=1$ . Hence, $R_{\mathcal {K}}(x,y) = t S_{n-1}(z) - S_{n-2}(z) = S_{n-1}(2) - S_{n-2}(2) = n-(n-1) =1 \not = 0$ .

If $S_{n-1}(z)=0$ , then $S^2_{n-1}(z) - z S_{n-1}(z) S_{n-2}(z) + S^2_{n-2}(z)=1$ (by Lemma 3.1) implies that $S^2_{n-2}(z) = 1$ . Thus, $R_{\mathcal {K}}(x,y) = t S_{n-1}(z) - S_{n-2}(z) = - S_{n-2}(z) = \pm 1 \not = 0$ .

Since $R_{\mathcal {K}}(x,y)=0$ for non-abelian $\mathrm {SL}_2(\mathbb C)$ -representations, we get $S_m(y) S_{n-1}(z) \not = 0$ .

$(2)$ Suppose $k=2m+1$ . For the holonomy representation $\rho _0$ , we have $x^2=4$ .

If $S_{m-1}(y)=0$ , then by Lemma 3.1, $y \in (-2, 2)$ . Moreover, $S^2_{m}(y) - y S_{m}(y) S_{m-1}(y) + S^2_{m-1}(y)=1$ implies that $S_m(y)= \pm 1$ . By (3.2) and (3.3), we have ${z = 2- (y-2) = 4-y \in (2, \infty )}$ and $t=1$ . Hence, $R_{\mathcal {K}}(x,y) = t S_{n-1}(z) - S_{n-2}(z) =S_{n-1}(z)-S_{n-2} (z)$ .

If $n=1$ , then $R_{\mathcal {K}}(x,y)=1$ . If $n \ge 2$ , then since $z \in (2, \infty )$ Lemma 3.1(3) implies that $R_{\mathcal {K}}(x,y) =S_{n-1}(z)-S_{n-2} (z) \not = 0$ . Similarly, if $n \le -1$ , then since $S_{-k}(z) = - S_{k-2}(z)$ we have $R_{\mathcal {K}}(x,y) =-S_{|n|-1}(z) + S_{|n|} (z) \not = 0$ .

If $S_m(y)=0$ , then by Lemma 3.1, $y \in (-2, 2)$ . Moreover, $S^2_{m}(y) - y S_{m}(y) S_{m-1}(y) + S^2_{m-1}(y)=1$ implies that $S_{m-1}(y)= \pm 1$ . By (3.2) and (3.3), we have $z=2- (y-2)= 4-y$ and $t=1-(y-2)=3-y$ . Hence, $t = z-1$ and

$$ \begin{align*}R_{\mathcal{K}}(x,y)= (z-1)S_{n-1}(z)-S_{n-2} (z)=S_n(z) - S_{n-1}(z). \end{align*} $$

As in the previous case, since $z \in (2, \infty )$ , we have $R_{\mathcal {K}}(x,y) \not =0$ . This contradicts the fact that $R_{\mathcal {K}}(x,y)=0$ for non-abelian $\mathrm {SL}_2(\mathbb C)$ -representations. Therefore, $S_m(y)S_{m-1}(y) \not = 0$ for the holonomy representation $\rho _0$ .

3.2 The twisted Alexander polynomial

Recall that the knot group of ${\mathcal {K}} = J(k, 2n)$ is $G({\mathcal {K}})= \langle a, b \mid w^n a = bw^n \rangle $ , where

$$ \begin{align*}w = \begin{cases} (ba^{-1})^m(b^{-1}a)^m& \text{if~} k=2m,\\ (ba^{-1})^mb(ab^{-1})^m a & \text{if~} k=2m+1. \end{cases} \end{align*} $$

Let $r = w^naw^{-n}b^{-1}$ . Then the adjoint twisted Alexander polynomial is given by $\Delta ^{\mathrm {Ad} \circ \rho }_{\mathcal {K}}(t) = \det \Phi \left ( \frac {\partial r}{\partial a} \right ) \big /\det \Phi (1-b)$ .

For $l \ge 0$ , we let $\sigma _l(u) = \sum _{i=0}^l u^i$ . We have

(3.4) $$ \begin{align} \frac{\partial r}{\partial a} &= \frac{\partial w^n}{\partial a} + w^n + w^n a \frac{\partial w^{-n}}{\partial a} \nonumber\\ &= \begin{cases} w^n \left( 1+ (1-a) w^{-n} \sigma_{n-1}(w) \frac{\partial w}{\partial a}\right) & \text{if~} n \ge 1,\\ w^{n} \left( 1 - (1-a) \sigma_{|n|-1}(w) \frac{\partial w}{\partial a}\right) & \text{if~} n \le -1. \end{cases} \end{align} $$

We now compute $\frac {\partial w}{\partial a}$ . If $k=2m$ , then $w =(ba^{-1})^m(b^{-1}a)^m$ and

$$ \begin{align*} \frac{\partial w}{\partial a} &= \frac{\partial (ba^{-1})^m}{\partial a} + (ba^{-1})^m \frac{\partial (b^{-1}a)^m}{\partial a} \\ &= -ba^{-1} \sigma_{m-1}(ba^{-1}) + (ba^{-1})^m b^{-1} \sigma_{m-1}(ab^{-1}). \end{align*} $$

If $k=2m+1 \ge 3$ , then $w = (ba^{-1})^mb(ab^{-1})^m a$ and

$$ \begin{align*} \frac{\partial w}{\partial a} &= \frac{\partial (ba^{-1})^m}{\partial a} + (ba^{-1})^m b \frac{\partial (ab^{-1})^m}{\partial a} + (ba^{-1})^mb(ab^{-1})^m \\ &= -ba^{-1} \sigma_{m-1}(ba^{-1}) + (ba^{-1})^m b \,\sigma_{m-1}(ab^{-1}) + (ba^{-1})^mb(ab^{-1})^m \\ &= -ba^{-1} \sigma_{m-1}(ba^{-1}) + (ba^{-1})^m b \,\sigma_{m}(ab^{-1}). \end{align*} $$

If $k=1$ , then $w=ba$ and $\frac {\partial w}{\partial a} = b$ . Hence,

(3.5) $$ \begin{align} \frac{\partial w}{\partial a} = \begin{cases} -ba^{-1} \sigma_{m-1}(ba^{-1}) + (ba^{-1})^m b^{-1} \sigma_{m-1}(ab^{-1}) & \text{if~} k=2m,\\ -ba^{-1} \sigma_{m-1}(ba^{-1}) + (ba^{-1})^m b \, \sigma_{m}(ab^{-1}) & \text{if~} k=2m+1 \ge 3, \\ b & \text{if~} k=1. \end{cases} \end{align} $$

Recall that the adjoint action $\mathrm {Ad}$ is the conjugation on $\mathrm {sl}_2(\mathbb C)$ by $\mathrm {SL}_2(\mathbb C)$ , that is, $\mathrm {Ad}(V)(g)=VgV^{-1} \in \mathrm {sl}_2(\mathbb C)$ for all $V \in \mathrm {SL}_2(\mathbb C)$ and $g \in \mathrm {sl}_2(\mathbb C)$ . We need the following lemma.

Lemma 3.5 [Reference Nguyen and Tran15, Lemma 2.1]

Suppose $V = \left [ \begin {array}{cc} e& f \\ g & h \end {array} \right ] \in \mathrm {SL}_2(\mathbb C)$ . Then

$$ \begin{align*}\mathrm{Ad}(V)=\left[ \begin{array}{ccc} e^2 & -2ef & -f^2 \\ -eg & eh + fg & fh \\ -g^2 & 2gh & h^2 \end{array} \right]. \end{align*} $$

Moreover, $\mathrm {Ad}(V) \in \mathrm {SL}_3(\mathbb C)$ .

Proposition 3.6 For $V \in \mathrm {SL}_2(\mathbb C)$ and $l \in \mathbb N$ , we have

$$ \begin{align*}\det \sigma_l \left( \mathrm{Ad}(V) \right) = (l+1) S^2_l(v),\end{align*} $$

where $v = {\mathrm {tr}} V$ .

Proof Up to conjugation, we can assume that $V= \left [ \begin {array}{cc} s& c \\ 0 & s^{-1} \end {array} \right ]$ for some $s \not = 0$ . Since $v = {\mathrm {tr}} V = s + s^{-1}$ , we have $S_l(v) = (s^{l+1} - s^{-l-1})/(s-s^{-1}) = s^{-l} (1 + s^2 + \cdots + s^{2l})$ .

By Lemma 3.5, we have $\mathrm {Ad}(V) =\left [ \begin {array}{ccc} s^2 & -2cs & -c^2 \\ 0 & 1& cs^{-1}\\ 0 & 0 & s^{-2} \end {array} \right ]$ . Then

$$ \begin{align*}\sigma_l (\mathrm{Ad}(V) ) = \sum_{i=0}^l (\mathrm{Ad}(V) )^i = \left[ \begin{array}{ccc} 1 + s^2 + \cdots + s^{2l} &* & * \\ 0 & l+1& *\\ 0 & 0 & 1 + s^{-2} + \cdots + s^{-2l} \end{array} \right]. \end{align*} $$

Hence,

$$ \begin{align*}\det \sigma_l (\mathrm{Ad}(V) ) = (l+1) s^{-2l} (1 + s^2 + \cdots + s^{2l})^2= (l+1)S^2_l(v).\\[-34pt] \end{align*} $$

We are ready to determine the highest and lowest degree terms of the adjoint twisted Alexander polynomial $\Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t)$ , which is equal to the adjoint hyperbolic torsion polynomial $\mathcal T^{\mathrm {Ad}}_{\mathcal {K}}(t)$ up to multiplication by $\pm t^{i}~(i\in {\mathbb Z})$ . Recall from (3.1) that, up to conjugation, a non-abelian representation $\rho : G({\mathcal {K}}) \to \mathrm {SL}_2(\mathbb C)$ has the form

$$ \begin{align*}\rho(a) = \left[ \begin{array}{cc} M & 1 \\ 0 & M^{-1} \end{array} \right] \quad \text{and} \quad \rho(b) = \left[ \begin{array}{cc} M & 0 \\ 2-y & M^{-1} \end{array} \right]. \end{align*} $$

Let $A=(\mathrm {Ad} \circ \rho ) (a)$ , $B=(\mathrm {Ad} \circ \rho ) (b)$ , and $W=(\mathrm {Ad} \circ \rho ) (w)$ , which are matrices in $\mathrm {SL}_3(\mathbb C)$ . Then $\Phi (a) = t A$ , $\Phi (b) = t B$ and $\Phi (w) = \begin {cases} t^0 W & \text {if~} k=2m,\\ t^6 W & \text {if~} k=2m+1. \end {cases}$

Since $\rho (b) = \left [ \begin {array}{cc} M & 0 \\ 2-y & M^{-1} \end {array} \right ]$ , by Lemma 3.5, we have

$$ \begin{align*}B = \mathrm{Ad}(\rho(b)) = \left[ \begin{array}{ccc} M^2 & 0 & 0 \\ -(2-y)M & 1 & 0 \\ -(2-y)^2 & 2(2-y)M^{-1} & M^{-2} \end{array} \right]. \end{align*} $$

Then $\det \Phi (1-b) = (1-t)(1 - tM^2)(1 - tM^{-2}) = (1-t)(t^2-t(x^2-2)+1)$ . Hence,

$$ \begin{align*}\Delta^{\mathrm{Ad} \circ \rho}_{\mathcal{K}}(t) = \frac{\det \Phi \left( \frac{\partial r}{\partial a} \right) }{(1-t)(t^2-t(x^2-2)+1)}. \end{align*} $$

For the holonomy representation $\rho _0$ , by [Reference Dubois and Yamaguchi2], the rational function $\Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t)$ becomes a Laurent polynomial in $t$ . Its highest and lowest degree terms are given as follows.

Proposition 3.7 If $n \ge 1$ , then

$$ \begin{align*} \Delta^{\mathrm{Ad} \circ \rho_0}_{\mathcal{K}}(t) = \begin{cases} -mn S^2_{m-1}(y) S^2_{n-1}(z) t^0 + \dots + mn S^2_{m-1}(y) S^2_{n-1}(z) t^{-3} & \text{if~} k=2m,\\ m S^2_{m-1}(y) t^{6n-3} + \dots - m S^2_{m-1}(y) t^{0} & \text{if~} k=2m+1 \ge 3, \\ - t^{6n-6} + \dots + t^{3} & \text{if~} k=1 \text{~and~} n \ge 2. \end{cases} \end{align*} $$

Proof Since $n \ge 1$ , by (3.4), we have $\frac {\partial r}{\partial a} = w^n \left ( 1+ (1-a) w^{-n} \sigma _{n-1}(w) \frac {\partial w}{\partial a}\right )$ .

If $k=2m$ , then $\frac {\partial w}{\partial a} = -ba^{-1} \sigma _{m-1}(ba^{-1}) + (ba^{-1})^m b^{-1} \sigma _{m-1}(ab^{-1})$ (by (3.5)). This implies that

$$ \begin{align*} \Phi \left( \frac{\partial r}{\partial a} \right) &= t A W^{-n} \sigma_{n-1}(W) BA^{-1} \sigma_{m-1}(BA^{-1}) \\ &\quad + \dots + t^{-1} W^{-n} \sigma_{n-1}(W) (BA^{-1})^m B^{-1} \sigma_{m-1}(AB^{-1}). \end{align*} $$

Taking determinant and applying Proposition 3.6, we obtain

$$ \begin{align*}\det \Phi \left( \frac{\partial r}{\partial a} \right) = t^3 n S^2_{n-1}(z) m S^2_{m-1}(y) + \dots + t^{-3} n S^2_{n-1}(z) m S^2_{m-1}(y). \end{align*} $$

Hence, $ \Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t) = - mn S^2_{m-1}(y) S^2_{n-1}(z) t^0 + \dots + mn S^2_{m-1}(y) S^2_{n-1}(z) t^{-3}$ .

If $k=2m+1 \ge 3$ , then by (3.5), we have $\frac {\partial w}{\partial a} = -ba^{-1} \sigma _{m-1}(ba^{-1}) + (ba^{-1})^m b \sigma _{m}(ab^{-1})$ . This implies that

$$ \begin{align*}\Phi \left( \frac{\partial r}{\partial a} \right) =t^{2n} ( I - AW^{-n} W^{n-1} (BA^{-1})^m B \, \sigma_{m}(AB^{-1}) ) + \dots - t^{0} W^{-n} BA^{-1} \sigma_{m-1}(BA^{-1})\end{align*} $$

Since

$$ \begin{align*} I - AW^{-n} W^{n-1} (BA^{-1})^m B \sigma_{m}(AB^{-1}) &= I - (AB^{-1})^{-m} \sigma_{m}(AB^{-1}) \\ &= I - \sigma_{m}(BA^{-1}) \\ &= - BA^{-1} \sigma_{m-1}(BA^{-1}), \end{align*} $$

we obtain

$$ \begin{align*}\Phi \left( \frac{\partial r}{\partial a} \right) = - t^{2n} BA^{-1} \sigma_{m-1}(BA^{-1}) + \dots - t^{0} W^{-n} BA^{-1} \sigma_{m-1}(BA^{-1}). \end{align*} $$

By Proposition 3.6, we have $\det \Phi \left ( \frac {\partial r}{\partial a} \right ) = - t^{6n} m S^2_{m-1}(y) + \dots - t^{0} m S^2_{m-1}(y)$ . Hence,

$$ \begin{align*}\Delta^{\mathrm{Ad} \circ \rho_0}_{\mathcal{K}}(t) = t^{6n-3} m S^2_{m-1}(y) + \dots - t^{0} m S^2_{m-1}(y). \end{align*} $$

If $k=1$ , then $w=ba$ and $\frac {\partial w}{\partial a} = b$ . We have

$$ \begin{align*}\Phi \left( \frac{\partial r}{\partial a} \right) = t^{2n} \left( I + (I - t A) (t^2BA)^{-n} \sigma_{n-1}(t^2 BA) (tB) \right). \end{align*} $$

Since $(tA) (t^2BA)^{-n} (t^2BA)^{n-1} (tB) = I$ , we get

$$ \begin{align*} \Phi \left( \frac{\partial r}{\partial a} \right) &= t^{2n}\left( (t^2BA)^{-n} \sigma_{n-1}(t^2 BA) (tB) - (t A) (t^2BA)^{-n} \sigma_{n-2}(t^2 BA) (tB) \right) \\ &= t^{2n} \left( (t^2BA)^{-n} (t^2BA)^{n-1} (tB) + \dots + (t^2BA)^{-n} (tB) \right). \end{align*} $$

Hence, $\det \Phi \left ( \frac {\partial r}{\partial a} \right ) = t^{6n-3} + \dots + t^3$ and $\Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t) = - t^{6n-6} + \dots +t^{3}$ .

Proposition 3.8 If $n \le -1$ , then

$$ \begin{align*} \Delta^{\mathrm{Ad} \circ \rho_0}_{\mathcal{K}}(t) = \begin{cases} m|n| S^2_{m-1}(y) S^2_{|n|-1}(z) t^0 + \dots - m|n| S^2_{m-1}(y) S^2_{|n|-1}(z) t^{-3} & \text{if~} k=2m,\\ - t^{-3} (m+1) S^2_{m}(y) + \dots +t^{6n}(m+1) S^2_{m}(y) & \text{if~} k=2m+1. \end{cases} \end{align*} $$

Proof Since $n \le -1$ , by (3.4), we have $\frac {\partial r}{\partial a} = w^{n} \left ( 1 - (1-a) \sigma _{|n|-1}(w) \frac {\partial w}{\partial a}\right )$ .

If $k=2m$ , then $\frac {\partial w}{\partial a} = -ba^{-1} \sigma _{m-1}(ba^{-1}) + (ba^{-1})^m b^{-1} \sigma _{m-1}(ab^{-1})$ . By Proposition 3.6, we have

$$ \begin{align*} \det \Phi \left( \frac{\partial r}{\partial a} \right) &= \det \left( - t A \sigma_{|n|-1}(W) BA^{-1} \sigma_{m-1}(BA^{-1}) \right) \\ & \quad + \dots + \det \left( - t^{-1} \sigma_{|n|-1}(W) (BA^{-1})^m B^{-1}\sigma_{m-1}(AB^{-1}) \right) \\ &= - t^3 |n| S^2_{|n|-1}(z) m S^2_{m-1}(y) + \dots - t^{-3} |n| S^2_{|n|-1}(z) m S^2_{m-1}(y). \end{align*} $$

Hence, $ \Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t) = m|n| S^2_{m-1}(y) S^2_{|n|-1}(z) t^0 + \dots - m|n| S^2_{m-1}(y) S^2_{|n|-1}(z) t^{-3}$ .

If $k=2m+1 \ge 3$ , then $\frac {\partial w}{\partial a} = -ba^{-1} \sigma _{m-1}(ba^{-1}) + (ba^{-1})^m b \, \sigma _{m}(ab^{-1})$ . By Proposition 3.6, we have

$$ \begin{align*} \det \Phi \left( \frac{\partial r}{\partial a} \right) &= t^{6n} \det \left( t^{2|n|} A W^{|n|-1} (BA^{-1})^m B \, \sigma_{m}(AB^{-1}) \right) \\ & \quad + \dots + t^{6n} \det \left( I + BA^{-1} \sigma_{m-1}(BA^{-1}) \right) \\ &= t^0 (m+1) S^2_{m}(y) + \dots + t^{6n}(m+1) S^2_{m}(y). \end{align*} $$

Note that $ I + BA^{-1} \sigma _{m-1}(BA^{-1}) = \sigma _{m}(BA^{-1})$ . Hence,

$$ \begin{align*}\Delta^{\mathrm{Ad} \circ \rho_0}_{\mathcal{K}}(t) = - t^{-3} (m+1) S^2_{m}(y) + \dots + t^{6n}(m+1) S^2_{m}(y).\end{align*} $$

If $k=1$ , then $w=ba$ and $\frac {\partial w}{\partial a} = b$ . We have

$$ \begin{align*}\det \Phi \left( \frac{\partial r}{\partial a} \right) = t^{6n} \det \left( t^{2|n|} A W^{|n|-1} B \right) + \dots + t^{6n} \det I = t^0 + \dots + t^{6n}. \end{align*} $$

Hence, $\Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t) = - t^{-3} + \dots + t^{6n}.$

3.3 Proof of Theorem 1.1

Recall that $\mathbb A$ denotes the set of all algebraic integers.

For the holonomy representation $\rho _0$ of a hyperbolic knot, we have ${\mathrm {tr}} \rho _0(\mu ) = \pm 2$ , where $\mu $ is any meridian. By [Reference Riley17], the Riley polynomial $R_K(\pm 2, y) \in \mathbb Z[y]$ of a two-bridge knot $K$ is monic. This implies that for the holonomy representation $\rho _0$ of a hyperbolic two-bridge knot, we have $y \in \mathbb A$ .

Consider the double twist knot ${\mathcal {K}} = J(k, 2n)$ and its holonomy representation $\rho _0$ . Since $x = \pm 2$ , by (3.2), we have

$$ \begin{align*}z = \begin{cases} 2 + (y-2)^2S^2_{m-1}(y) & \text{if~} k=2m,\\ 2 - (y-2) \left( S_m(y) - S_{m-1}(y) \right)^2 & \text{if~} k=2m+1. \end{cases} \end{align*} $$

The proof of Theorem 1.1 is divided into three cases: $(1)\, k=2m$ , $(2)\, k=2m+1 \ge 3$ , and $(3)\, k=1$ .

Case 1: $k=2m$ . Note that $J(2m, 2n)$ has genus one. Moreover, $J(2m, 2n)$ is fibered if and only if $m=|n|=1$ .

Since $S_{-n-1}(z) = - S_{n-1}(z)$ , by Proposition 3.4, we have $S^2_{m-1}(y) S^2_{|n|-1}(z) \not = 0$ . Hence, Propositions 3.7 and 3.8 imply that the adjoint twisted Alexander polynomial $\Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t)$ is a polynomial of degree $3$ with highest coefficient $- mn S^2_{m-1}(y) S^2_{|n|-1}(z)$ .

If $m|n|=1$ , then $m|n| S^2_{m-1}(y) S^2_{|n|-1}(z)=1$ . Hence, $\Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t)$ is monic.

If $m|n| \ge 2$ , then by applying Corollary 2.2 with $q=m|n|$ , $r = \pm 1$ , and $P(y) = S^2_{m-1}(y) S^2_{|n|-1}(z) \in \mathbb Z[y]$ and noting that $y \in \mathbb A$ , we obtain $m|n| S^2_{m-1}(y) S^2_{|n|-1}(z) \pm 1 \not = 0$ . This means that $\Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t)$ is not monic.

Case 2: $k=2m+1 \ge 3$ . Note that $J(2m+1, 2n)$ has genus $|n|$ . Moreover, $J(2m+1, 2n)$ is fibered if and only if $m=1$ and $n \ge 1$ .

By Proposition 3.4, we have $S_m(y) \not = 0$ and $S_{m-1}(y) \not = 0$ .

If $n \ge 1$ , then by Proposition 3.7, $\Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t)$ is a polynomial of degree $6n-3$ with highest coefficient $mS^2_{m-1}(y)$ . If $m=1$ , then $mS^2_{m-1}(y)=1$ and hence $\Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t)$ is monic. If $m \ge 2$ , then by applying Corollary 2.2 with $q=m$ , $r = \pm 1$ , and $P(y) = S^2_{m-1}(y) \in \mathbb Z[y]$ and noting that $y \in \mathbb A$ , we obtain $mS^2_{m-1}(y) \pm 1 \not = 0$ . Thus, $\Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t)$ is not monic.

If $n \le -1$ , then by Proposition 3.8, $\Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t)$ is a polynomial of degree ${-3+6|n|}$ with highest coefficient $-(m+1)S^2_{m}(y)$ . Since $y \in \mathbb A$ , Corollary 2.2 implies that ${(m+1)S^2_{m}(y) \pm 1 \not = 0}$ . This means that $\Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t)$ is not monic.

Case 3: $k=1$ . Note that $J(1, 2n)$ is fibered and its genus is given by

$$ \begin{align*}g(J(1, 2n)) = \begin{cases} n-1 & \text{if~} n \ge 1,\\ |n| & \text{if~} n \le -1. \end{cases} \end{align*} $$

If $n \ge 2$ , then by Proposition 3.7, $\Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t)$ is a monic polynomial of degree $6n-9$ .

If $n \le -1$ , then by Proposition 3.8, $\Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t)$ is a monic polynomial of degree ${-3+6|n|}$ .

In all three cases shown, we have $\deg \Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}} (t) = 3 (2 g({\mathcal {K}})-1)$ . Moreover, $\Delta ^{\mathrm {Ad} \circ \rho _0}_{\mathcal {K}}(t)$ is monic if and only if ${\mathcal {K}}$ is fibered. This completes the proof of Theorem 1.1.